Group Action of Symmetric Group/Subset

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $\N_n$ denote the set $\set {1, 2, \ldots, n}$.

Let $\struct {S_n, \circ}$ denote the symmetric group on $\N_n$.


Let $r \in \N: 0 < r \le n$.

Let $B_r$ denote the set of all subsets of $\N_n$ of cardinality $r$:

$B_r := \set {S \subseteq \N_n: \card S = r}$


Let $*$ be the mapping $*: S_n \times B_r \to B_r$ defined as:

$\forall \pi \in S_n, \forall S \in B_r: \pi * B_r = \pi \sqbrk S$

where $\pi \sqbrk S$ denotes the image of $S$ under $\pi$.


Then $*$ is a group action.


Proof

The group action axioms are investigated in turn.

Let $\pi, \rho \in S_n$.

Let $S \in B_r$.


Thus:

\(\ds \pi * \paren {\rho * S}\) \(=\) \(\ds \pi * \rho \sqbrk S\) Definition of $*$
\(\ds \) \(=\) \(\ds \pi \sqbrk {\rho \sqbrk S}\) Definition of $*$
\(\ds \) \(=\) \(\ds \paren {\pi \circ \rho} \sqbrk S\) Definition of Composition of Mappings
\(\ds \) \(=\) \(\ds \paren {\pi \circ \rho} * n\) Definition of $*$

demonstrating that Group Action Axiom $\text {GA} 1$ holds.


Then:

\(\ds I_{\N_n} * S\) \(=\) \(\ds I_{\N_n} \sqbrk S\) where $I_{\N_n}$ is the identity mapping on $\N_n$
\(\ds \) \(=\) \(\ds S\) Definition of Identity Mapping

demonstrating that Group Action Axiom $\text {GA} 2$ holds.

$\blacksquare$


Sources